tag:blogger.com,1999:blog-4575155025214649471.post2716216598240189746..comments2024-03-20T00:19:58.164-07:00Comments on CBSE Pathshala: ASSIGNMENT, 2012-13 Class : IX, Sub : Maths TOPIC : Heron’s Formula & Co-ordinate GeometryJ.Sunil Sirhttp://www.blogger.com/profile/06753061330031546203noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-4575155025214649471.post-91577855071023532142012-10-22T23:26:39.848-07:002012-10-22T23:26:39.848-07:00Find the area of quadrilateral ABCD whose sides ar...Find the area of quadrilateral ABCD whose sides are 9m,40m.28m and 15m is.................<br />Let AB =9 m, BC=40m, CD= 28m and DA = 15m<br />In quad. ABCD, join diagonal AC<br />Take a look at the lengths of the sides given closely. You will see that 92 + 402 = 412. (Pythgorean triplets) So in trainagle ABC, AB and BC are two sides of a right triangle with Angle ABC =90 deg with its diagonal being AC. Apply pythagoras theorem for triangle ABC to get AC = 41cm.<br />Now we find two triangles ABC and ACD with sides already known. Apply Heron 's formula to get the areas of the two triangles. (Alternately, area of triangle ABC can be found more easily since BC is the height and AB is the base.)<br />Area of ABC = 180 cm2. Area of ACD = 126 cm2.<br />Simply sum the two areas to get the total area to be 306 cm2.<br />J.Sunil Sirhttps://www.blogger.com/profile/06753061330031546203noreply@blogger.comtag:blogger.com,1999:blog-4575155025214649471.post-8678171942141472842012-10-16T06:21:16.994-07:002012-10-16T06:21:16.994-07:00Very informative post indeed.. being enrolled in: ...Very informative post indeed.. being enrolled in: http://www.wiziq.com/course/7618-full-preparation-for-class-10-mathematics<br />I was looking for such articles online to assist me and also your post helped me a lot.:)Anonymoushttps://www.blogger.com/profile/10873886598653528045noreply@blogger.com